Mixing
Irreducible components of the subvariety of matrices s.t. $M^n=mathbb{I}$
$begingroup$
Fixed $ninmathbb{N}$, describe the irreducible components, and prove that there are ${n+1choose{2}}$ of them, of the following algebraic variety:
$$V={Ain M_{2}(mathbb{C})mid A^n=mathbb{I}}$$
For $n=2$, the components are ${mathbb{I}}$, ${-mathbb{I}}$ and
$$ left{ begin{pmatrix} a & b\ c & d end{pmatrix}: a=-d , a^2+bc=1 right}$$
For the other values of $n$, I don't have idea how to proceed
algebraic-geometry affine-varieties
asked Jan 6 at 5:16
jonlajoyejonlajoye
13
$endgroup$
add a comment |
$begingroup$
Fixed $ninmathbb{N}$, describe the irreducible components, and prove that there are ${n+1choose{2}}$ of them, of the following algebraic variety:
$$V={Ain M_{2}(mathbb{C})mid A^n=mathbb{I}}$$
For $n=2$, the components are ${mathbb{I}}$, ${-mathbb{I}}$ and
$$ left{ begin{pmatrix} a & b\ c & d end{pmatrix}: a=-d , a^2+bc=1 right}$$
For the other values of $n$, I don't have idea how to proceed
algebraic-geometry affine-varieties
asked Jan 6 at 5:16
jonlajoyejonlajoye
13
$endgroup$
$begingroup$
So you have for $n=2$ three components: these are distinguished by their eigenvalues, respectively $(1,1)$, $(-1,-1)$ and $(1,-1)$, and therefore also by their characteristic equations.
$endgroup$
– Lord Shark the Unknown
Jan 6 at 5:37
$begingroup$
Ok, so the minimal polinomial of A must divide $x^n-1$. Let $a_{1}$...$a_{n}$ be the roots of the unity. Then, if the minimal polinomial is linear, we obtain $A=a_imathbb{I}$, and thus $P_A(x)=(x-a_i)^2$ . Otherwise, in order to buld a possible characteristical polynomial we have to chose two differents roots, no matter the order. In total, we obtain $(n+1chose2)$ different polynomials. Each of one determines an irreducible component. If I am right I will edit the answer
$endgroup$
– jonlajoye
Jan 6 at 6:06
add a comment |
$begingroup$
Fixed $ninmathbb{N}$, describe the irreducible components, and prove that there are ${n+1choose{2}}$ of them, of the following algebraic variety:
$$V={Ain M_{2}(mathbb{C})mid A^n=mathbb{I}}$$
For $n=2$, the components are ${mathbb{I}}$, ${-mathbb{I}}$ and
$$ left{ begin{pmatrix} a & b\ c & d end{pmatrix}: a=-d , a^2+bc=1 right}$$
For the other values of $n$, I don't have idea how to proceed
algebraic-geometry affine-varieties
asked Jan 6 at 5:16
jonlajoyejonlajoye
13
$endgroup$
Fixed $ninmathbb{N}$, describe the irreducible components, and prove that there are ${n+1choose{2}}$ of them, of the following algebraic variety:
$$V={Ain M_{2}(mathbb{C})mid A^n=mathbb{I}}$$
For $n=2$, the components are ${mathbb{I}}$, ${-mathbb{I}}$ and
$$ left{ begin{pmatrix} a & b\ c & d end{pmatrix}: a=-d , a^2+bc=1 right}$$
For the other values of $n$, I don't have idea how to proceed
algebraic-geometry affine-varieties
algebraic-geometry affine-varieties
asked Jan 6 at 5:16
jonlajoyejonlajoye
13
asked Jan 6 at 5:16
jonlajoyejonlajoye
13
edited Jan 6 at 5:33
jonlajoye
asked Jan 6 at 5:16
jonlajoyejonlajoye
13
asked Jan 6 at 5:16
jonlajoyejonlajoye
13
asked Jan 6 at 5:16
jonlajoyejonlajoye
13
13
$begingroup$
So you have for $n=2$ three components: these are distinguished by their eigenvalues, respectively $(1,1)$, $(-1,-1)$ and $(1,-1)$, and therefore also by their characteristic equations.
$endgroup$
– Lord Shark the Unknown
Jan 6 at 5:37
$begingroup$
Ok, so the minimal polinomial of A must divide $x^n-1$. Let $a_{1}$...$a_{n}$ be the roots of the unity. Then, if the minimal polinomial is linear, we obtain $A=a_imathbb{I}$, and thus $P_A(x)=(x-a_i)^2$ . Otherwise, in order to buld a possible characteristical polynomial we have to chose two differents roots, no matter the order. In total, we obtain $(n+1chose2)$ different polynomials. Each of one determines an irreducible component. If I am right I will edit the answer
$endgroup$
– jonlajoye
Jan 6 at 6:06
add a comment |
$begingroup$
So you have for $n=2$ three components: these are distinguished by their eigenvalues, respectively $(1,1)$, $(-1,-1)$ and $(1,-1)$, and therefore also by their characteristic equations.
$endgroup$
– Lord Shark the Unknown
Jan 6 at 5:37
$begingroup$
Ok, so the minimal polinomial of A must divide $x^n-1$. Let $a_{1}$...$a_{n}$ be the roots of the unity. Then, if the minimal polinomial is linear, we obtain $A=a_imathbb{I}$, and thus $P_A(x)=(x-a_i)^2$ . Otherwise, in order to buld a possible characteristical polynomial we have to chose two differents roots, no matter the order. In total, we obtain $(n+1chose2)$ different polynomials. Each of one determines an irreducible component. If I am right I will edit the answer
$endgroup$
– jonlajoye
Jan 6 at 6:06
$begingroup$
So you have for $n=2$ three components: these are distinguished by their eigenvalues, respectively $(1,1)$, $(-1,-1)$ and $(1,-1)$, and therefore also by their characteristic equations.
$endgroup$
– Lord Shark the Unknown
Jan 6 at 5:37
$begingroup$
So you have for $n=2$ three components: these are distinguished by their eigenvalues, respectively $(1,1)$, $(-1,-1)$ and $(1,-1)$, and therefore also by their characteristic equations.
$endgroup$
– Lord Shark the Unknown
Jan 6 at 5:37
$begingroup$
Ok, so the minimal polinomial of A must divide $x^n-1$. Let $a_{1}$...$a_{n}$ be the roots of the unity. Then, if the minimal polinomial is linear, we obtain $A=a_imathbb{I}$, and thus $P_A(x)=(x-a_i)^2$ . Otherwise, in order to buld a possible characteristical polynomial we have to chose two differents roots, no matter the order. In total, we obtain $(n+1chose2)$ different polynomials. Each of one determines an irreducible component. If I am right I will edit the answer
$endgroup$
– jonlajoye
Jan 6 at 6:06
$begingroup$
Ok, so the minimal polinomial of A must divide $x^n-1$. Let $a_{1}$...$a_{n}$ be the roots of the unity. Then, if the minimal polinomial is linear, we obtain $A=a_imathbb{I}$, and thus $P_A(x)=(x-a_i)^2$ . Otherwise, in order to buld a possible characteristical polynomial we have to chose two differents roots, no matter the order. In total, we obtain $(n+1chose2)$ different polynomials. Each of one determines an irreducible component. If I am right I will edit the answer
$endgroup$
– jonlajoye
Jan 6 at 6:06
add a comment |
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$begingroup$
So you have for $n=2$ three components: these are distinguished by their eigenvalues, respectively $(1,1)$, $(-1,-1)$ and $(1,-1)$, and therefore also by their characteristic equations.
$endgroup$
– Lord Shark the Unknown
Jan 6 at 5:37
$begingroup$
Ok, so the minimal polinomial of A must divide $x^n-1$. Let $a_{1}$...$a_{n}$ be the roots of the unity. Then, if the minimal polinomial is linear, we obtain $A=a_imathbb{I}$, and thus $P_A(x)=(x-a_i)^2$ . Otherwise, in order to buld a possible characteristical polynomial we have to chose two differents roots, no matter the order. In total, we obtain $(n+1chose2)$ different polynomials. Each of one determines an irreducible component. If I am right I will edit the answer
$endgroup$
– jonlajoye
Jan 6 at 6:06